(20y-11)+(4y+6)(7y-7)=360

Solution for (20y-11)+(4y+6)(7y-7)=360 equation:

(20y-11)+(4y+6)(7y-7)=360

We move all terms to the left:

(20y-11)+(4y+6)(7y-7)-(360)=0

We get rid of parentheses

20y+(4y+6)(7y-7)-11-360=0

We multiply parentheses ..

(+28y^2-28y+42y-42)+20y-11-360=0

We add all the numbers together, and all the variables

(+28y^2-28y+42y-42)+20y-371=0

We get rid of parentheses

28y^2-28y+42y+20y-42-371=0

We add all the numbers together, and all the variables

28y^2+34y-413=0

a = 28; b = 34; c = -413;
Δ = b2-4ac
Δ = 342-4·28·(-413)
Δ = 47412
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{47412}=\sqrt{36*1317}=\sqrt{36}*\sqrt{1317}=6\sqrt{1317}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-6\sqrt{1317}}{2*28}=\frac{-34-6\sqrt{1317}}{56}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+6\sqrt{1317}}{2*28}=\frac{-34+6\sqrt{1317}}{56}
$